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Differentiation

 

A. Derivative Using First Principle /ab Initio Method

 

If f(x) is a derivable function then, Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.1 Using first principles, find the derivative of the function y =-cot x- x.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.2 Find by first principle the derivative of Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE  w.r.t. x.

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.3 If f (x) = (ln x)x , find f ' (x) from the first principle.

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Derivative Of Standard Functions :

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

B. Rules of Differentiation

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE i.e. multiply the differential coefficient of each separate function by the product of all the remaining functions and add up all the results; the sum will be the differential coefficient of the product of all the functions.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Ex.4 Differentiate y = sin(x2)

 

Sol. If y = sin(x2), then the outer function is the sine function and the inner function is the squaring function, so the Chain Rule gives

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.5 If F(x) = (x2 + 2)3, compute F'(x). One way to do this problem is to expand (x2 + 2)3 and use the differentiation formulae.

 

Sol. F(x) = (x2 + 2)3 = x6 + 6x4 + 12x2 + 8,

F'(x) = 6x5 + 24x3 + 24x.

Another method uses the Chain Rule. Let g and f be the functions defined, respectively, by g(x) = x2 + 2 and f(y) = y3. Then f(g(x)) = (x2 + 2)3 = F(x), and, according to the Chain Rule,

F'(x) = [f(g(x))]' = f'(g(x)) g'(x).

Since g'(x) = 2x and f'(y) = 3y2, we get f'(g(x)) = 3(x2 + 2)2 and

F'(x) = 3(x2 + 2)2 (2x) = 6x (x4 + 4x2 + 4),

which agrees with the alternative solution above.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Let f and g be two differentiable functions. The formation of the composite function f(g) is suggested by writing u = g(x) and y = f(u). Thus x is transformed by g into u, and the resulting u is then transformed by f into y = f(u) = f(g(x)). We have

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

The idea that one can simply cancel out du in (1) is very appealing and accounts for the popularity of the notation. It is important to realize that the cancellation is valid because the Chain Rule is true, and not vice versa. Thus, du is simply a part of the notation for the derivative and means nothing by itself. Note also that (1) is incomplete in the sense that it does not say explicitly at what points to evaluate the derivatives. We can add this information by writing

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.6  Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol. 

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.7 Differentiate g(x) = Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol.  Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE is the outer function.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.8 If y = Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE , find Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE .

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.9   Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

C. Logarithmic Differentiation

To find the derivative of :

(i) a function which is the product or quotient of a number of functions OR

(ii) a function of the form [f(x)]g(x) where f & g are both derivable, it will be found convenient to take the logarithm of the function first & then differentiate. This is called Logarithmic Differentiation.

Steps in Logarithmic Differentiation

1. Take natural logarithms of both sides of an equation y = f(x) and use the Laws of Logarithms to simplify.

2. Differentiate both sides with respect to x.

3. Solve the resulting equation for y'.

If f(x) < 0 for some values of x, then in f(x) is not defined, but we can write |y| = |f(x)| and

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE
 

Ex.10 Find

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol. Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.11 Differentiate y = Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE .

 

Sol. Using logarithmic differentiation, we have Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.12 Find Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE , where y = (x + 1)2x.

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

is the sum of the two special forms and therefore we may, instead of taking logarithms in any particular example, consider first u constant and then v constant and add the results obtained on these suppositions.

 

D. Parametric Differentiation

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

E. Differentiation of Implicit Functions

Assume that the equation F(x, y) = 0 specifies y as an implicit function of x. In what follows we shall consider this function to be differentiable.

Differentiating both sides of the equation F(x, y) = 0 with respect to x, we obtain a first-degree equation with respect to y'. This equation easily yields y', that is, the derivative of the implicit function.

 

Ex.13 Find Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE from the equation x3 + ln y- x2ey = 0.

 

Sol. Differentiating both sides of the equation with respect to x, we obtain

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Steps for Implicit Differentiation

1. Differentiate both sides of the equation with respect to x.

2. Collect all terms involving dy/dx on the left side of the equation and move all other terms to the right side of the equation.

3. Factor dy/dx out of the left side of the equation.

4. Solve for dy/dx by dividing both sides of the equation by the left-hand factor that does not contain dy/dx.

 

Ex.14 Find dy/dx given that y3 + y2- 5y- x2 =-4.

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.15 Find Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE if xy- x = 1.

 

Sol. We must find Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE at x = 1. Assume y is a function of x,y = y(x). The relation now is xy (x)- x = 1.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE
 

Remark If, in the preceding example, we had explicitly solved for y, we would have obtained

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

To see that this is exactly the result obtained by implicit methods, note that if 1 + 1/x is substituted for y ⇒ Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.16 Find dy/dx if 2x2 + xy- 3y2 = x.

 

Sol. If we assume y is a function of x and represent it by y(x), the equation reads 2x2 + xy(x)- 3[y(x)]2 = x.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Be careful : When applying Formula (2), keep in mind that the only values of x and y that can be substituted into the right-hand side of Formula (2) are those values that satisfy the original condition 2x2 + xy- 3y2 = x. for instance, we might substitute x = 1, y = 2 to obtain (dy/dx) = (1- 4- 2)/(1- 12) = 5/11; however, (1, 2) is not a point on the graph 2x2 + xy- 3y2 = x, so the calculation of dy/dx at this point is totally meaningless.

The equation in Example can be written as -3y2 + xy + (2x2- x) = 0 and hence is a quadratic equation in y (an equation of the form Ay2 + By + C = 0 where A =-3, B = x and C = 2x2- x). Hence, we could use the quadratic formula to solve this equation for y in terms of x, obtaining

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Though we are able to find y explicitly in terms of x, the resulting expression is fairly complex, and it still might be best to find dy/dx implicitly as in Example.

Warning : It is important to realize that implicit differentiation is a technique for finding dy/dx that is valid only if y is a differentiable function of x, and careless application of the technique can lead to errors. For example, there is clearly no real-valued function y = f(x) that satisfies the equation
x2 + y2 =-1, yet formal application of implicit differentiation yields the :derivative: dy/dx =-x/y. To be able to evaluate this "derivative," we must find some values for which x2 + y2 =-1. Because no such values exist, the derivative does not exist.

 

Ex.17 If x3 + y2 = sin (x + y), find Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol. given, x3 + y3 = sin (x + y) ...(i)

differentiating w. r. t. x, we get Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.18 If x = Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE , prove that Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE = 2x2 + y2- 3xy

 

Sol.

x = Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

F. Derivative of Inverse Functions

If the inverse functions f & g are defined by y = f(x) & x = g(y) & if

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

The truth of this is also manifested geometrically, for Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE and Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE are respectively the tangent and the co-tangent of the angle y which the tangent to the curve y = f(x) makes with the x-axis.

This formula is very useful in the differentiation of an inverse function.

 

Ex.19 Find the derivative of Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE  and mention the points of non differentiability.

 

Sol. Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

f is continuous for all x but not diff. at x = 1 ,-1

 

G. Derivative of a Determinant

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.20 Let f(x) = Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE . Show that f '' (x) = 0 and that f(x) = f(0) + k x where k denotes the sum of all the co-factors of the elements in f(0).

 

Sol. 

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

H. Higher order Derivatives

Let a function y = f(x) be defined on an open interval (a, b) . It's derivative, if it exists on (a, b) is a certain function f '(x) [or (dy/dx) or y ] & is called the first derivative of y w. r. t. x.

If it happens that the first derivative has a derivative on (a , b) then this derivative is called the second derivative of y w. r. t. x & is denoted by f ''(x) or (d2y/dx2) or y '' .

Once we have found the derivative f' of any function f, we can go on and find the derivative of f'.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Similarly, the 3rd order derivative of y w. r. t. x , if it exists, is defined by Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE .

It is also denoted by f '''(x) or y '''.

The third derivative, written f''', is the derivative of the second derivative, and, in principle, we can go on forever and form derivatives of higher order. We adopt the alternative notation f(n) for the nth derivative of f.

 

Ex.21 Find y'' if x4 + y4 = 16.

 

Sol. Differentiating the equation implicitly with respect to x, we get 4x3 + 4y3y' = 0

Solving for y' gives y' = -Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

to find y'' we differentiate this expression for y' using the Quotient Rule and remembering that y is a function of Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

If we now substitute Equation 1 into this expression, we get

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

But the values of x and y must satisfy the original equation x4 + y4 = 16. So the answer simplifies to

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.22 If y = x sin x, prove that  Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE = 0

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.23 A function f(x) is so defined that for al x, [f(x)]n = f(nx). Prove that f(x) . f'(nx) = f'(x) . f(nx), where f'(x) denotes derivative of f(x) w.r. to x.

 

Sol. Given [f(x)]n = f(nx)                                                                 .....(1)

differentiating both sides w. r to x, we get

n[f(x)]n- 1 . f'(x) = f'(nx) . (n . 1)             or,                   [f(x)n . f'(x) = f'(nx)

multiplying both sides by f'(x), we get

[f(x)n . f'(x) = f'(nx) . f(x) or, f(nx) . f'(x) = f'(nx) . f(x)                           [from (1)]

 

Ex.24 y =  Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE , then prove that, Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE = x2 + 4.

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

I. L' Hopital's Rule

Let f and g be functions that are differentiable on an open interval (a, b) containing c, except possibly at c itself. Assume that g'(x) Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE 0 for all x in (a, b), except possibly at c itself. If the limit of f(x)/g(x) as x approaches c produces the indeterminate frorm 0/0, then

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

provided the limit on the right exists (or is infinite). This result also applies if the limit of f(x)/g(x) as x approaches c produces any one of the indeterminate forms

∞ / ∞,     (- ∞) / ∞   , ∞ / (- ∞).

 

Ex.25   Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE          Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE         L' Hopital's Rule.

 

Sol. Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE.

 

Ex.26 Determine Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.27 Find Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol. The inconvenience of continuously differentiating the denominator, which involves tan2 x as a factor, may be partially avoided as follows. We write

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

To evaluate the limit on the R.H.S., we notice that the numerator and denominator both become 0 for x = 0.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.28 Find the constants 'a' (a > 0) and 'b' such that Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE = 1.

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.29 Evaluate Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Ex.30 Evaluate : Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

 

Sol.

Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

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FAQs on Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE

1. What is differentiation and what are its applications?
Ans. Differentiation is the process of calculating the rate of change of a function with respect to one of its variables. It is used in many fields such as physics, engineering, and economics to analyze and model relationships between variables. Some of its applications include finding maximum and minimum values of functions, determining rates of change, and analyzing the behavior of functions at various points.
2. What are the basic rules of differentiation?
Ans. The basic rules of differentiation include the power rule, product rule, quotient rule, chain rule, and sum and difference rule. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1). The product rule states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). The quotient rule states that if f(x) = u(x)/v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)]/v^2(x). The chain rule states that if f(g(x)) is a composite function, then f'(g(x)) = f'(u)*u'(x), where u = g(x). The sum and difference rule states that if f(x) = u(x) + v(x), then f'(x) = u'(x) + v'(x).
3. How do you find the derivative of a function using differentiation?
Ans. To find the derivative of a function, we use the rules of differentiation to calculate the rate of change of the function with respect to one of its variables. For example, if we have the function f(x) = x^2 + 3x, we can use the power rule to find that f'(x) = 2x + 3, which gives us the rate of change of the function at any point x.
4. What is the difference between differentiation and integration?
Ans. Differentiation and integration are inverse operations of each other. Differentiation calculates the rate of change of a function with respect to one of its variables, while integration calculates the area under the curve of a function. In other words, differentiation finds the slope of the tangent line to a curve at a given point, while integration finds the area between the curve and the x-axis.
5. What is the importance of differentiation in calculus?
Ans. Differentiation is one of the fundamental concepts in calculus and is used to model and analyze relationships between variables. It is important in many fields such as physics, engineering, and economics to determine rates of change, find maximum and minimum values of functions, and analyze the behavior of functions at various points. Differentiation also forms the basis for many other concepts in calculus such as optimization, related rates, and curve sketching.
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